Problem 59 - Calculus Early Transcendentals

Question

Find the volume of the following solids using the method of your choice. The solid whose base is the region bounded by \(y=x^{2}\) and the line \(y=1\) and whose cross sections perpendicular to the base and parallel to the \(x\) -axis are semicircles

Step-by-Step Solution

Answer

Answer: The volume of the solid is \(\frac{16\pi}{15}\).

Step 1: Determine the Range of Integration

To determine the range of integration, we first need to find the x values at which the parabola \(y=x^2\) intersects the line \(y=1\). We can solve for these intersection points by setting the equations equal to each other and solve for x: \(x^2 = 1\) \(x=\pm1\) So, the base of the solid lies between \(x=-1\) and \(x=1\). This will be our range of integration.

Step 2: Find the Area of Each Cross-sectional Semicircle

The cross-sections are semicircles, so we will find the area of each semicircle based on the length of the diameter (which is parallel to the x-axis). The diameter will equal the distance between the parabola and the line \(y=1\) for each x value. This distance can be found by subtracting the y value of the parabola, \(x^2\), from the y value of the line: Diameter = \(1-x^2\) Since the area of a circle is given by \(A_{circle}=\pi r^2\), and the area of a semicircle is half the area of a circle, we can find the area of a semicircle with radius r as follows: \(A_{semicircle}=\frac{1}{2}\pi r^2\) The radius of the semicircle is half the diameter, so: Radius = \(\frac{1-x^2}{2}\) Now we can substitute this expression for the radius into the semicircle area formula: \(A_{semicircle}=\frac{1}{2}\pi\left(\frac{1-x^2}{2}\right)^2\)

Step 3: Integrate to Find the Volume

Now that we have the area of a semicircle for each value of x, we can integrate this expression over the range of \(x=-1\) to \(x=1\) to find the volume of the solid. We have: \(Volume=\int_{-1}^{1}\frac{1}{2}\pi\left(\frac{1-x^2}{2}\right)^2dx\) This integral can be solved either by substitution or by expanding the expression and integrating term by term: \(Volume=\int_{-1}^{1}\left(\frac{\pi}{4}(1-2x^2+x^4)\right)dx\) Using the power rule and the fundamental theorem of calculus, we have: \(Volume=\frac{\pi}{4}\left[x-\frac{2}{3}x^3+\frac{1}{5}x^5\right]_{-1}^{1}\) Evaluating the integral at the limits, we find: \(Volume=\frac{\pi}{4}\left[(1-\frac{2}{3}+\frac{1}{5})-(-1+\frac{2}{3}-\frac{1}{5})\right]=\frac{16\pi}{15}\) Therefore, the volume of the solid is \(\frac{16\pi}{15}\).

Key Concepts

Intersection Points

To find the volume of a solid, understanding the region over which you calculate is essential. The first step is identifying where the curves intersect, as these points determine the boundaries of the solid.
The given exercise involves a parabola, expressed as \(y = x^2\), and a horizontal line, \(y = 1\). To locate their points of intersection, set these two equations equal:

• Set \(x^2 = 1\).
• Solve for \(x\) to find \(x= \pm 1\).

These calculations show that the parabola intersects the line at \(x = -1\) and \(x = 1\). These points become the limits or range of integration for the problem, setting the stage for subsequent calculations.

Semicircle Cross-Sections

With the intersection points known, the next task is to determine the characteristics of the cross-sections of the solid. In this problem, each cross-section is a semicircle.
The first consideration is that the semicircles are perpendicular to the base along the x-axis. Therefore, their diameter can be described as the vertical distance between the parabola \(y=x^2\) and the line \(y=1\) at each value of \(x\), computed as:

• Diameter = \(1-x^2\).

Knowing the diameter, you can find the radius by dividing the diameter by 2:

• Radius = \(\frac{1-x^2}{2}\).

The formula for the area of a semicircle uses this radius. Since the area of a full circle is \(\pi r^2\), for a semicircle, the area is:

• \(A_{semicircle} = \frac{1}{2} \pi \left( \frac{1-x^2}{2} \right)^2\).

This expression offers the area of each individual semicircular cross-section that will be considered in finding the complete volume of the solid.

Integration

The final step in computing the volume of a solid is integration. Integration, a core calculus concept, measures the accumulation of quantities; in this case, the accumulated areas of semicircular cross-sections along the defined limits.
With the radius and area of each semicircle known, set up the integral over the range \(-1 \le x \le 1\):

• The volume integral expression is:
• \[Volume = \int_{-1}^{1} \frac{1}{2} \pi \left(\frac{1-x^2}{2}\right)^2 \, dx\].

To solve this integral, simplify first:

• \[Volume = \int_{-1}^{1} \left( \frac{\pi}{4} (1 - 2x^2 + x^4 ) \right) \, dx\].

Apply the power rule and evaluate the integral:

• \[Volume = \frac{\pi}{4} \left[ x - \frac{2}{3}x^3 + \frac{1}{5}x^5 \right]_{-1}^{1}\].

Substituting the limits of integration in gives:

• \[Volume = \frac{16\pi}{15}\].

This calculation completes the task, providing the exact volume of the solid whose base and structure were defined by the initial intersection points and the geometry of the semicircles.